Molar mass KCl = 74.55 g/mol
Number of moles:
mass KC / molar mass
149 / 74.55 => 1.998 moles
Volume in liters: 500 mL / 1000 => 0.5 L
Therefore:
M = moles / volume
M = 1.998 / 0.5
M = 3.996 mol/L⁻¹
Answer:
2.16 × 10⁻³
Explanation:
Step 1: Given data
Concentration of the acid (Ca): 0.260 M
Acid dissociation constant (Ka): 1.80 × 10⁻⁵
Step 2: Write the acid dissociation equation
HC₂H₃O₂(aq) + H₂O(l) ⇄ C₂H₃O₂⁻(aq) + H₃O⁺(aq)
Step 3: Calculate the concentration of H₃O⁺ at equilibrium
We will use the following expression.
![[H_3O^{+} ]= \sqrt{Ka \times Ca } = \sqrt{1.80 \times 10^{-5} \times 0.260 } = 2.16 \times 10^{-3}](https://tex.z-dn.net/?f=%5BH_3O%5E%7B%2B%7D%20%5D%3D%20%5Csqrt%7BKa%20%5Ctimes%20Ca%20%7D%20%3D%20%5Csqrt%7B1.80%20%5Ctimes%2010%5E%7B-5%7D%20%5Ctimes%200.260%20%7D%20%3D%202.16%20%5Ctimes%2010%5E%7B-3%7D)
Answer:
2MnO2 (s)
Explanation:
3KCN(aq) + 2KMnO4(aq) + 1H2O(l) → 3KCNO(aq) + 2MnO2(s) + 2KOH(aq)
Reactants:
K: 5
C: 3
N: 3
Mn: 2
O: 9
H: 2
Products:
K: 5
C: 3
N: 3
Mn: 2
O: 9
H: 2
It depends on where it is locacated (close or far away from the equator)